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4 years ago

Ecuaciones cuadraticas 4x²- 20x=0 3x²-42=0 2x²-14x=0 8x²-98=0

Mathematics

1 answer:

Setler79 [48]4 years ago

3 0

Im so sorry.. I did not mean to submit an answer..

And I honestly don't know how to get rid of my answer.. but I will try to help.

4x²- 20x=0 = $x_{1}$ = -√5, $x_{2}$ = √5

3x²-42=0 = $x_{1}$ = -√14, $x_{2}$ = √14

2x²-14x=0 = $x_{1}$ = 0, $x_{2}$ = 7

8x²-98=0 = $x_{1}$ = - $\frac{7}{2}$ , $x_{2}$ = $\frac{7}{2}$

I hope that these are the answers that you wanted, if they are not, I am sorry, tell me and I will try to fix them.

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3 years ago

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Answer:

Step-by-step explanation:

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3 years ago

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3 years ago

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Read 2 more answers

Solve to find x log (3x+5) to base 5= 2log (1-3x) to base 25

Eduardwww [97]

Answer:

x = -2/ 3

Step-by-step explanation:

in order to cancel out the logs they should have common bases

$\mathsf{ log_{5}(3x + 5) = 2 log_{25}(1 - 3x) }$

we can write 25 as 5²

$\mathsf{ \implies log_{5}(3x + 5) = 2 \: log_{ {5}^{2} }(1 - 3x) }$

we know that the reciprocal of the exponents of the bases are multiplied to the log

$\mathsf{ \implies log_{5}(3x + 5) = \frac{1}{\cancel{2} } \times \cancel{2} \: log_{ 5}(1 - 3x) }$

and now since the logs have common bases

$\mathsf{ \implies \cancel { log_{5}}(3x + 5) = \cancel{ log_{ 5}}(1 - 3x) }$

we're left with

$\mathsf{ \implies 3x + 5 = 1 - 3x}$

$\mathsf{ \implies 9x = -4}$

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3 years ago